Termination w.r.t. Q of the given QTRS could not be shown:

0 QTRS
↳1 Overlay + Local Confluence (⇔)
↳2 QTRS
↳3 DependencyPairsProof (⇔)
↳4 QDP
↳5 DependencyGraphProof (⇔)
↳6 AND
  ↳7 QDP
    ↳8 QDPOrderProof (⇔)
    ↳9 QDP
    ↳10 PisEmptyProof (⇔)
    ↳11 TRUE
  ↳12 QDP
    ↳13 QDPOrderProof (⇔)
    ↳14 QDP
    ↳15 DependencyGraphProof (⇔)
    ↳16 TRUE
  ↳17 QDP
    ↳18 QDPOrderProof (⇔)
    ↳19 QDP
    ↳20 PisEmptyProof (⇔)
    ↳21 TRUE
  ↳22 QDP

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

last(nil) → 0
last(cons(x, nil)) → x
last(cons(x, cons(y, xs))) → last(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
reverse(nil) → nil
reverse(cons(x, xs)) → cons(last(cons(x, xs)), reverse(del(last(cons(x, xs)), cons(x, xs))))

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

last(nil) → 0
last(cons(x, nil)) → x
last(cons(x, cons(y, xs))) → last(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
reverse(nil) → nil
reverse(cons(x, xs)) → cons(last(cons(x, xs)), reverse(del(last(cons(x, xs)), cons(x, xs))))

The set Q consists of the following terms:

last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
reverse(nil)
reverse(cons(x0, x1))

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

Q DP problem:
The TRS P consists of the following rules:

LAST(cons(x, cons(y, xs))) → LAST(cons(y, xs))
DEL(x, cons(y, xs)) → IF(eq(x, y), x, y, xs)
DEL(x, cons(y, xs)) → EQ(x, y)
IF(false, x, y, xs) → DEL(x, xs)
EQ(s(x), s(y)) → EQ(x, y)
REVERSE(cons(x, xs)) → LAST(cons(x, xs))
REVERSE(cons(x, xs)) → REVERSE(del(last(cons(x, xs)), cons(x, xs)))
REVERSE(cons(x, xs)) → DEL(last(cons(x, xs)), cons(x, xs))

The TRS R consists of the following rules:

last(nil) → 0
last(cons(x, nil)) → x
last(cons(x, cons(y, xs))) → last(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
reverse(nil) → nil
reverse(cons(x, xs)) → cons(last(cons(x, xs)), reverse(del(last(cons(x, xs)), cons(x, xs))))

The set Q consists of the following terms:

last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
reverse(nil)
reverse(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 3 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

Q DP problem:
The TRS P consists of the following rules:

EQ(s(x), s(y)) → EQ(x, y)

The TRS R consists of the following rules:

last(nil) → 0
last(cons(x, nil)) → x
last(cons(x, cons(y, xs))) → last(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
reverse(nil) → nil
reverse(cons(x, xs)) → cons(last(cons(x, xs)), reverse(del(last(cons(x, xs)), cons(x, xs))))

The set Q consists of the following terms:

last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
reverse(nil)
reverse(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


EQ(s(x), s(y)) → EQ(x, y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
EQ(x1, x2)  =  x2
s(x1)  =  s(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

last(nil) → 0
last(cons(x, nil)) → x
last(cons(x, cons(y, xs))) → last(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
reverse(nil) → nil
reverse(cons(x, xs)) → cons(last(cons(x, xs)), reverse(del(last(cons(x, xs)), cons(x, xs))))

The set Q consists of the following terms:

last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
reverse(nil)
reverse(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.

(10) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(11) TRUE

(12) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF(false, x, y, xs) → DEL(x, xs)
DEL(x, cons(y, xs)) → IF(eq(x, y), x, y, xs)

The TRS R consists of the following rules:

last(nil) → 0
last(cons(x, nil)) → x
last(cons(x, cons(y, xs))) → last(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
reverse(nil) → nil
reverse(cons(x, xs)) → cons(last(cons(x, xs)), reverse(del(last(cons(x, xs)), cons(x, xs))))

The set Q consists of the following terms:

last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
reverse(nil)
reverse(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


DEL(x, cons(y, xs)) → IF(eq(x, y), x, y, xs)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
IF(x1, x2, x3, x4)  =  x4
false  =  false
DEL(x1, x2)  =  x2
cons(x1, x2)  =  cons(x2)
eq(x1, x2)  =  x1
s(x1)  =  s
0  =  0
true  =  true

Lexicographic Path Order [LPO].
Precedence:
s > [false, 0]


The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF(false, x, y, xs) → DEL(x, xs)

The TRS R consists of the following rules:

last(nil) → 0
last(cons(x, nil)) → x
last(cons(x, cons(y, xs))) → last(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
reverse(nil) → nil
reverse(cons(x, xs)) → cons(last(cons(x, xs)), reverse(del(last(cons(x, xs)), cons(x, xs))))

The set Q consists of the following terms:

last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
reverse(nil)
reverse(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.

(15) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(16) TRUE

(17) Obligation:

Q DP problem:
The TRS P consists of the following rules:

LAST(cons(x, cons(y, xs))) → LAST(cons(y, xs))

The TRS R consists of the following rules:

last(nil) → 0
last(cons(x, nil)) → x
last(cons(x, cons(y, xs))) → last(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
reverse(nil) → nil
reverse(cons(x, xs)) → cons(last(cons(x, xs)), reverse(del(last(cons(x, xs)), cons(x, xs))))

The set Q consists of the following terms:

last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
reverse(nil)
reverse(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.

(18) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


LAST(cons(x, cons(y, xs))) → LAST(cons(y, xs))
The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic Path Order [LPO].
Precedence:
[LAST1, cons2]


The following usable rules [FROCOS05] were oriented: none

(19) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

last(nil) → 0
last(cons(x, nil)) → x
last(cons(x, cons(y, xs))) → last(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
reverse(nil) → nil
reverse(cons(x, xs)) → cons(last(cons(x, xs)), reverse(del(last(cons(x, xs)), cons(x, xs))))

The set Q consists of the following terms:

last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
reverse(nil)
reverse(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.

(20) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(21) TRUE

(22) Obligation:

Q DP problem:
The TRS P consists of the following rules:

REVERSE(cons(x, xs)) → REVERSE(del(last(cons(x, xs)), cons(x, xs)))

The TRS R consists of the following rules:

last(nil) → 0
last(cons(x, nil)) → x
last(cons(x, cons(y, xs))) → last(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
reverse(nil) → nil
reverse(cons(x, xs)) → cons(last(cons(x, xs)), reverse(del(last(cons(x, xs)), cons(x, xs))))

The set Q consists of the following terms:

last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
reverse(nil)
reverse(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.